Answer to Question #97370 in Statistics and Probability for Tiffany

Question

1) Forty percent of the english students at Parkway High school had Dr.Mima for history, and the rest had Dr. P. Thirty percent of Dr.Mima's students earn an A in probability, while only ten percent of Dr. P's students earn an A in probability.

a) What percent of Parkway High School history students earn an A?

A= have Dr.Mima=0.40

B= rest have Dr.P=0.60

C= Dr.Mima students earn A = 0.30

C= Dr.P students earn A = 0.10



P(C)=P(A)P(C|A) + P(B)P(C|B)

= (0.40)(0.30) + (0.60)(10)

P(C) = 0.19 -answer



b) Tiffany did not earn an A in english. What is the probability that Tiffany had Dr.P for history?

Answer 1

a) What percent of Parkway High School history students earn an A?

A= have Dr.Mima=0.40

B= rest have Dr.P=0.60

C= student earns A

$P(C|A)=0.3$

$P(C|B)=0.1$

Total probability

$$P(C)=P(C|A)P(A)+P(C|B)P(B)$$ $$P(C)=0.3(0.4)+0.6(0.1)=0.18$$

18% of Parkway High School history students earn an A.

b) Tiffany did not earn an A in probability. What is the probability that Tiffany had Dr.P for history?

Let C be the event that student earns an A in probability. Then

$$P(\bar{C}|A)=1-0.3=0.7$$

$$P(\bar{C}|B)=1-0.1=0.9$$

Use the Bayes' rule

$$P(B|\bar{C})={P(\bar{C}|B)P(B) \over P(\bar{C}|B)P(B)+P(\bar{C}|A)P(A)}$$

$$P(B|\bar{C})={0.9(0.6) \over 0.9(0.6)+0.7(0.4)}={27 \over 41}\approx0.66$$

The probability that Tiffany had Dr.P for history is ${27 \over 41}\approx0.66$